pub fn new_birthday_probability(n: u32) -> f64 {
     if n > 365 {
        return 1.0;
    }

    // 计算所有人生日都不相同的概率
    let mut prob = 1.0;
    for i in 0..n {
        prob *= (365 - i) as f64 / 365.0;
    }

    // 至少两个人生日相同的概率
    1.0 - prob
}
